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Question 14 - NCERT Exemplar - MCQs - Chapter 6 Class 12 Application of Derivatives (Term 1)

Last updated at March 22, 2023 by

The tangent to the curve γ€–y=eγ€—^2x at the point (0, 1) meets x-axis at:

(A)(0, 1)Β  Β  Β  Β  Β  (B) βˆ’ 1/2, 0

(C) (2, 0)Β  Β  Β  Β  Β (D) (0, 2)

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Transcript

Question 14 The tangent to the curve 〖𝑦=𝑒〗^2π‘₯ at the point (0, 1) meets x-axis at: (0, 1) (B) βˆ’ 1/2, 0 (C) (2, 0) (D) (0, 2) Given 〖𝑦=𝑒〗^2π‘₯ Finding slope of tangent at (0, 1) 〖𝑦=𝑒〗^2π‘₯ Differentiating w.r.t. π‘₯ π’…π’š/𝒅𝒙 = 2𝒆^πŸπ’™ At point (0, 1) Putting π‘₯=0, 𝑦=0 in 𝑑𝑦/𝑑π‘₯ 𝑑𝑦/𝑑π‘₯ = γ€–2𝑒〗^0 𝑑𝑦/𝑑π‘₯ = 2 Γ— 1 π’…π’š/𝒅𝒙=𝟐 Finding equation of tangent Equation of line at (π‘₯1 , 𝑦1) & having Slope m is π‘¦βˆ’π‘¦1=π‘š(π‘₯βˆ’π‘₯1) ∴ Equation of tangent at point (0, 1) & having slope 2 is (π’šβˆ’πŸ) = 2 (π’™βˆ’πŸŽ) (π‘¦βˆ’1) = 2π‘₯ π’š=πŸπ’™+𝟏 Now, we need to find the point where tangent meets the x-axis Since point is on the x-axis, it’s y-coordinate will be 0 ∴ Point = (𝒙,𝟎) Putting 𝑦 = 0 in (1) 0 = 2𝒙 + 1 βˆ’1 = 2π‘₯ (βˆ’1)/2=π‘₯ 𝒙 = (βˆ’πŸ)/𝟐 Thus, required point is ((βˆ’πŸ)/𝟐, 𝟎) So, the Correct answer is (B)

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